查询daabase失败未选择数据库

时间:2018-09-24 13:21:04

标签: php mysql

我遇到错误,没有选择任何数据库,任何人都可以更正代码。

我正在尝试学习php和mysql。所以我尝试使用
建立数据库   phpmyadmin并将其与我的php连接。这是一个简单的示例,在这里我尝试查看数据库是否正常工作。

    <?php
    //Get values passe from form in donateform.html.
        $link;
        $Name = $_POST['Name'];
        $Mobile = $_POST['Mobile'];
        $Email = $_POST['Email'];
        $Donating =isset($_POST['Donating']);
        $Address = isset($_POST['Address']);
    //To prevent mysql injection
        $Name = stripcslashes($Name);
        $Mobile = stripcslashes($Mobile);
        $Email = stripcslashes($Email);
        $Donating = stripcslashes($Donating);
        $Address = stripcslashes($Address);

    //connect to the server and select  database.
        $link=mysqli_connect("localhost", "gooddeeds", "");
        mysqli_select_db($link,"donaters");

        $Name = mysqli_real_escape_string($link,$Name);
        $Mobile = mysqli_real_escape_string($link,$Mobile);
        $Email = mysqli_real_escape_string($link,$Email);
        $Donating = mysqli_real_escape_string($link,$Donating);
        $Address = mysqli_real_escape_string($link,$Address);

//Query the database for user
$result =mysqli_query($link,"INSERT INTO donaters (Name, Mobile, Email, Donating, Address) VALUES ('$Name', '$Mobile', '$Email', '$Donating', '$Address')")
    or die("Failed to query daabase".mysqli_error($link));

if(mysqli_query($result)){
    echo "Records inserted successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. ".mysqli_error($link);
}
?>

我得到

Database query failed::: No database selected

这意味着这部分代码

//connect to the server and select  database.
    $link=mysqli_connect("localhost", "gooddeeds", "");
    mysqli_select_db($link,"donaters");

不起作用(如果有,我会在其中分别添加不同数量的“:”。将不胜感激!谢谢!

2 个答案:

答案 0 :(得分:0)

您可以尝试使用以下作为databasename.tablename(gooddeeds.donaters)

INSERT INTO gooddeeds.donaters (Name, Mobile, Email, Donating, Address) VALUES ('$Name', '$Mobile', '$Email', '$Donating', '$Address')

答案 1 :(得分:0)

这是php文档中的第一个示例。尝试使用条件检查连接。

$link = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db");

if (!$link) {  
    echo "Error: Unable to connect to MySQL." . PHP_EOL;  
    echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    exit;
}